ZOJ 2974 Just Pour the Water

矩阵快速幂。

构造一个矩阵，$a[i][j]$表示一次操作后，$j$会从$i$那里得到水的比例。注意$k=0$的时候，要将$a[i][j]$置为$1$。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0);
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = ;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c))
    {
        x = x *  + c - '';
        c = getchar();
    }
}
 
struct Matrix
{
    double A[][];
    int R, C;
    Matrix operator*(Matrix b);
};
 
Matrix X, Y, Z;
 
int T,n,p;
 
Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, , sizeof(c.A));
    int i, j, k;
    for (i = ; i <= R; i++)
        for (j = ; j <= b.C; j++)
            for (k = ; k <= C; k++)
                c.A[i][j] = c.A[i][j] + A[i][k] * b.A[k][j];
    c.R = R; c.C = b.C;
    return c;
}
 
void init()
{
    Y.R = n; Y.C = n;
    for (int i = ; i <= n; i++) Y.A[i][i] = ;
    X.R = n; X.C = n;
    Z.R = ; Z.C = n;
}
 
void work()
{
    while (p)
    {
        if (p %  == ) Y = Y*X;
        p = p >> ;
        X = X*X;
    }
    Z = Z*Y;
}
 
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
 
        memset(X.A, , sizeof X.A);
        memset(Y.A, , sizeof Y.A);
        memset(Z.A, , sizeof Z.A);
 
        for(int i=;i<=n;i++) scanf("%lf",&Z.A[][i]);
 
        for(int i=;i<=n;i++)
        {
            int K; scanf("%d",&K);
            for(int j=;j<=K;j++)
            {
                int id; scanf("%d",&id);
                X.A[i][id]=1.0/K;
            }
            if(K==)
            {
                X.A[i][i]=1.0;
            }
        }
 
        scanf("%d",&p);
 
        init();
        work();
 
        for(int i=;i<=n;i++)
        {
            printf("%.2f",Z.A[][i]);
            if(i<n) printf(" "); else printf("\n");
        }
    }
    return ;
}